Problem: The lifespans of zebras in a particular zoo are normally distributed. The average zebra lives $20.3$ years; the standard deviation is $3.1$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a zebra living longer than $23.4$ years.
Explanation: $20.3$ $17.2$ $23.4$ $14.1$ $26.5$ $11$ $29.6$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $20.3$ years. We know the standard deviation is $3.1$ years, so one standard deviation below the mean is $17.2$ years and one standard deviation above the mean is $23.4$ years. Two standard deviations below the mean is $14.1$ years and two standard deviations above the mean is $26.5$ years. Three standard deviations below the mean is $11$ years and three standard deviations above the mean is $29.6$ years. We are interested in the probability of a zebra living longer than $23.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the zebras will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the zebras will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $17.2$ years and the other half $({16\%})$ will live longer than $23.4$ years. The probability of a particular zebra living longer than $23.4$ years is ${16\%}$.